Answer:
a) 
b) 
Explanation:
Given data:
Electric field = 1.47 N/C
velocity of electron is 
distance of point b from point A is 0.55 m
we know that acceleration of particle is given as
a) for electron



from equation of motion we have



b) for proton


from equation of motion we have



Answer:
net power is + 2.25 D
Explanation:
Given data
distance vision = -0.25 D
near vision = + 2.50 D
to find out
net power
solution
we have given a person lens power for near is - 0.25 diopter and lens power for near power is +2.50 diopter so
net power is sum of both the power vision
so
net power = distance + near power
put both value we get net power
net power = ( -0.25 D) + ( + 2.50 D)
net power = + 2.25 D
so net power is + 2.25 D
Answer:
The approximate terminal velocity of a sky diver before the parachute opens is 320 km/h.
Explanation:
- The terminal velocity is the maximum magnitude of velocity that is attained by the diver when he or she falls in the air.
- The terminal velocity of the person diving in air before opening parachute is 320 km/h that means the velocity when the person is experiencing free fall is 320 km/h.
- During terminal velocity, we can represent mathematical equation as;
Buoyancy force + drag force = Gravity
When a river flows into an ocean, it slows down and deposits materials in its delta